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Difference between revisions of "2011 AMC 8 Problems/Problem 16"
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<cmath> b = 2x</cmath>
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<cmath>\boxed{\textbf{(C) } a = b} </cmath>
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<cmath>\boxed{\textbf{(C) } A = B} </cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=15|num-a=17}}
 
{{AMC8 box|year=2011|num-b=15|num-a=17}}
 
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{{MAA Notice}}

Revision as of 14:51, 18 October 2016

Problem

Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$?

$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad\textbf{(D) } A = \dfrac43B \\ \\ \textbf{(E) }A = \dfrac{16}9B$

Solution

25-25-30

We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have \[15^2 + x^2 =25^2\] \[x^2 = 25^2 - 15^2\] \[x^2 = (25 + 15)(25-15)\] \[x^2= 40\cdot 10\] \[x^2= 400\] \[x = \sqrt{400}\] \[x= 20\]


Thus we have two 15-20-25 right triangles.

25-25-40

We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles. Let the area of a 15-20-25 right triangle be $x$. \[a = 2x\] \[b = 2x\] \[\boxed{\textbf{(C) } A = B}\]

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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